1.56 mm
0.78 mm
2.34 mm
0.90 mm
Let the temperature of the junction be T.
∴rate of heat transfer = dQ/dt = 2KA(T - 10)/L = KA(400 - T)/L
⇒ 2(T - 10) = 400 - T
or, T = 140 ℃
Now, for the wire PQ, let us imagine a small length Δx at a distance x from the junction.
∴ ΔT/Δx = (140 - 10)/1 = 130
So, temperature at distance x: T = 10 + 130x
or, T - 10 = 130x
Increase in length of the small element Δx is expressed as:
dy/dx = αΔT = α(T - 10)
or, dy/dx = α × 130x
Integrating both sides, we get:
∫₀ΔL dy = 130α∫₀¹ x dx
∵ L = 1 m (Given)
or, ΔL = 130αx²/2 = 130 × 1.2 × 10⁻⁵ × 1²/2 = 0.78 × 10⁻³ m = 0.78 mm